Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 14:50

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# One black, one red, one blue and two white pearls are

Author Message
Manager
Joined: 07 Nov 2004
Posts: 89
Location: London
Followers: 1

Kudos [?]: 1 [0], given: 0

One black, one red, one blue and two white pearls are [#permalink]

### Show Tags

07 Nov 2004, 15:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

One black, one red, one blue and two white pearls are attached to make a necklace. What is the probability that the black and the blue pearls are next to each other?
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4289
Followers: 43

Kudos [?]: 469 [0], given: 0

### Show Tags

07 Nov 2004, 15:59
Got 1/2 here
Total possible outcome: (n-1)! / 2 --> Because we are talking about beads on a necklace, we can just "flip" it over and we have the exact same reverse pattern so we have to eliminate those "mirror" patterns. 4!/2 = 12
You then have to divide it further by 2 because there are 2 white pearls. 12/2 = 6
I then just fixed the black pearl drew out the possible outcomes when the blue pearl is next to it and there are 3 possible ways to fix them next to each other.
3/6 = 1/2
Took me too long for a problem which might seem simple at first.
_________________

Best Regards,

Paul

Manager
Joined: 07 Nov 2004
Posts: 89
Location: London
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

07 Nov 2004, 16:43
OA=1/2
Took me ages too.
Intern
Joined: 12 Aug 2004
Posts: 33
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

07 Nov 2004, 17:32
I'm not sure why you need to divide by two because there are two white pearls, how does this change the fact that there is a total of 5 pearls, does it mater if 1,2 or three are white, all that matters is the there is 1 out of 5 that is black and 1 out of 5 that is blue, right?
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4289
Followers: 43

Kudos [?]: 469 [0], given: 0

### Show Tags

07 Nov 2004, 17:59
jeremy02 wrote:
I'm not sure why you need to divide by two because there are two white pearls, how does this change the fact that there is a total of 5 pearls, does it mater if 1,2 or three are white, all that matters is the there is 1 out of 5 that is black and 1 out of 5 that is blue, right?

Below you will see a diagram which will explain what I mean. In a ring like formation, there are (n-1)! ways of arranging the components by fixing 1 item down. In the drawing, let's say that the black pearl is anchored, there are 4! ways of arranging the remaining items. However, because there are 2 white pearls, you can "interchange them" and you will have to further reduce by 2! the possible number of outcomes. Now, we are down to 4!/2! = 12.

The next concept is a bit more subtle. Normally, when we are fixing people around a round table, (n-1)! would suffice to give you the answer. However, we are talking about beads on a necklace. Once you have one formation, you can "flip" it over and you will get the "mirror" image. Look carefully at the diagram and you will see that although the two necklaces are arranged differently, the right-hand side one is just the "mirror" image of the left-hand side one. Hence, you have to divide everything by another factor of 2 because of the possibility of making another outcome merely by flipping the necklace over. 12/2 = 6
Attachments

GMATClub3.jpg [ 6.68 KiB | Viewed 755 times ]

_________________

Best Regards,

Paul

Manager
Joined: 28 Jul 2004
Posts: 54
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

07 Nov 2004, 21:17
i did this in a diff way...

first the black ball is picked.... then the porbablity that the blue ball will be picked is 1/4

but there are two ways of doing this so...

I got the answer as 1/4 * 2 =1/2

is this a right way of doin it ?
_________________

Jim

Manager
Joined: 28 Aug 2004
Posts: 205
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

15 Nov 2004, 07:53
total possible outcome is (n-1)! / 2 = 4!/2 = 12.

Black and Blue next to each other (BB):

1. WBBR
2. RBBW
3. BBRW
4. BBWR
5. WRBB
6. RWBB

BB could be in the middle; to the left; or to the right (relatively speaking).

Pr = favorable outcome / total outcome = 6/12 = 1/2.
Director
Joined: 31 Aug 2004
Posts: 607
Followers: 3

Kudos [?]: 132 [0], given: 0

### Show Tags

15 Nov 2004, 09:56
This one is nice ! From where it is Oxon ?
Manager
Joined: 12 May 2004
Posts: 126
Followers: 1

Kudos [?]: 85 [0], given: 0

### Show Tags

16 Nov 2004, 02:36
My explanation is:

No of ways to arrange n pearls in a necklace: (n-1)! Because two pearls have same color, so total ways are (n-1)!/2 = (5-1)!/2 = A

If blue and black stands together, so we can consider combination of blue and black pearls as one set. Now we have 4 pearls (3 individual and 1 set). Ways to arrange 4 those pearls are (4-1)!/2 (reasoning is the same as above).

Addtionally, within the set we have two ways to turn 2 pearls around. Hence total of possible ways to arrange those pearls are

2*(4-1)!/2 = B

Probability is B/A = 0.5
Manager
Joined: 07 Nov 2004
Posts: 89
Location: London
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

16 Nov 2004, 14:53
twixt,
Got it from an old A-level text book.
16 Nov 2004, 14:53
Display posts from previous: Sort by